Edexcel International GCSE Mathematics A 4MA1: Simultaneous linear equations – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.6 Simultaneous linear equations in Topic 2 of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions marked (Both tiers) test statement 2.6A, which is on Foundation and Higher papers. Questions and parts marked (Higher) test statement 2.6B, interpreting the equations as lines, which is Higher tier only. A calculator may be used on every 4MA1 paper, but show your working: most of the marks here are for method.

Learn the content first in the study guide and the revision notes. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement.

Questions

1. (Both tiers) Solve the simultaneous equations x + y = 17 and x − y = 5. [2]

2. (Both tiers) Solve the simultaneous equations 3x + 2y = 18 and x + 2y = 10. [3]

3. (Both tiers) Solve the simultaneous equations 4p − q = 11 and 3p + 2q = 22. [3]

4. (Both tiers) Solve the simultaneous equations y = 4x − 5 and 3x + 2y = 23. [3]

5. (Both tiers) In a café, 2 coffees and 3 muffins cost £13.20. 4 coffees and 1 muffin cost £14.40. A coffee costs £c and a muffin costs £m.

(a) Write down two equations in c and m. [2] (b) Calculate the cost of one coffee and the cost of one muffin. [3]

6. (Both tiers) Solve the simultaneous equations 5x + 3y = 1 and 2x − 7y = 25. Show clear algebraic working. [4]

7. (Both tiers) Solve the simultaneous equations 6x − 4y = 1 and 3x + 8y = 13. Give your answers as exact values. [3]

8. (Both tiers) Solve the simultaneous equations x/2 + y/3 = 4 and 3x − y = 6. [4]

9.

(a) Solve the simultaneous equations y = 3x − 4 and 2x + y = 11. [3] (b) (Higher) The lines y = 3x − 4 and 2x + y = 11 are drawn on the same grid. Write down the coordinates of the point where they cross. [1] (c) (Higher) Explain why the simultaneous equations y = 3x − 4 and 6x − 2y = 5 have no solution. [2]

10. (Both tiers) The sides of a rectangle, in centimetres, are labelled as follows: top (5x − y), bottom (3x + 7), left (2y − 1) and right (x + 3).

(a) Use the fact that opposite sides of a rectangle are equal to write two equations. Simplify each to the form ax + by = c. [2] (b) Solve your equations to find x and y. [3] (c) Calculate the perimeter of the rectangle. [1] (d) Calculate the area of the rectangle. [1]

11. (Both tiers) A taxi company charges a fixed amount of £F plus £r for every kilometre travelled. A journey of 8 km costs £14.60. A journey of 13 km costs £21.60.

(a) Write down two equations in F and r. [2] (b) Find the values of F and r. [3] (c) Calculate the cost of a journey of 20 km. [1] (d) The cost, £C, of a journey of d km is C = F + rd. The graph of C against d is a straight line. State what the gradient and the C-intercept of this line represent. [2]

12. Here are two equations: kx + 3y = 7 and 2x − y = 7, where k is a constant.

(a) When k = 4, solve the equations. Give your answers as exact values. [3] (b) (Higher) Find the value of k for which the equations have no solution. Explain your answer by referring to the lines the equations represent. [3]

Answers

1. Adding the equations: 2x = 22 [1]; x = 11, y = 6 [1] Examiner insight: Both values are needed for the final mark; x = 11 on its own earns only the method mark.

2. Subtracting: (3x + 2y) − (x + 2y) = 18 − 10, so 2x = 8 [1]; x = 4 [1]; 4 + 2y = 10, y = 3 [1] Examiner insight: Adding the equations here gives 4x + 4y = 28, which removes nothing; the method mark needs a correct elimination of one letter.

3. Second equation unchanged, first × 2: 8p − 2q = 22; adding: 11p = 44 [1]; p = 4 [1]; 16 − q = 11, q = 5 [1] Examiner insight: The method mark is for multiplying correctly and then adding or subtracting so that one letter goes; if one term is not multiplied, the method mark is lost and the answers usually fail too.

4. 3x + 2(4x − 5) = 23 [1]; 3x + 8x − 10 = 23, 11x = 33, x = 3 [1]; y = 4(3) − 5, y = 7 [1] Examiner insight: Writing 3x + 8x − 5 = 23 (only part of the bracket multiplied) leads to x = 28/11; check that both terms inside the bracket were multiplied.

5. (a) 2c + 3m = 13.20 [1]; 4c + m = 14.40 [1] (b) First equation × 2: 4c + 6m = 26.40; subtracting the second: 5m = 12.00 [1]; m = 2.40 [1]; 4c + 2.40 = 14.40, c = 3.00 [1]. A coffee costs £3.00 and a muffin costs £2.40. Examiner insight: The accuracy marks need both prices correct and clearly matched to the right item; write money to 2 decimal places (£2.40) and say which value is the coffee and which is the muffin.

6. First × 2 and second × 5: 10x + 6y = 2 and 10x − 35y = 125 [1]; subtracting: 41y = −123 [1]; y = −3 [1]; 5x − 9 = 1, x = 2 [1] Examiner insight: The second method mark is for a correct subtraction; 6y − (−35y) must give 41y, and an error such as −29y loses that mark and both accuracy marks.

7. First × 2: 12x − 8y = 2; adding the second: 15x = 15 [1]; x = 1 [1]; 3 + 8y = 13, 8y = 10, y = 5/4 [1] Examiner insight: 5/4 and 1.25 are both exact and both acceptable; a rounded value such as 1.3 would not be.

8. Multiply the first equation by 6: 3x + 2y = 24 [1]; subtracting 3x − y = 6: 3y = 18 [1]; y = 6 [1]; 3x − 6 = 6, x = 4 [1] Examiner insight: The first method mark is for clearing the fractions with every term multiplied; multiplying x/2 and y/3 by 6 but leaving the 4 unchanged gives 3x + 2y = 4 and loses all four marks.

9. (a) 2x + (3x − 4) = 11 [1]; 5x = 15, x = 3 [1]; y = 5 [1] (b) (3, 5) [1] (c) 6x − 2y = 5 rearranges to y = 3x − 2.5 [1]; both lines have gradient 3 but different y-intercepts (−4 and −2.5), so they are parallel and never meet, so there is no point that satisfies both equations [1] Examiner insight: In (c) “the lines are parallel” needs a reason; the marks come from showing the equal gradients after rearranging, not from the claim alone.

10. (a) 5x − y = 3x + 7 gives 2x − y = 7 [1]; 2y − 1 = x + 3 gives x − 2y = −4 [1] (b) First equation × 2: 4x − 2y = 14; subtracting x − 2y = −4: 3x = 18 [1]; x = 6 [1]; 12 − y = 7, y = 5 [1] (c) Sides are 5(6) − 5 = 25 cm and 2(5) − 1 = 9 cm, so perimeter = 2(25 + 9) = 68 cm [1] (d) 25 × 9 = 225 cm² [1] Examiner insight: Marks in (c) and (d) can follow through from wrong values of x and y, but only if the side lengths are worked out from the expressions; a check that 5x − y and 3x + 7 give the same length catches errors early.

11. (a) F + 8r = 14.60 [1]; F + 13r = 21.60 [1] (b) Subtracting: 5r = 7.00 [1]; r = 1.40 [1]; F = 14.60 − 8(1.40) = 14.60 − 11.20, F = 3.40 [1] (c) 3.40 + 20(1.40) = 3.40 + 28.00 = £31.40 [1] (d) The gradient, 1.40, is the cost per kilometre (£1.40 per km) [1]; the C-intercept, 3.40, is the fixed charge (£3.40), paid before any distance is travelled [1] Examiner insight: In (d) each mark needs the meaning in context; “the gradient is 1.40” without saying it is the charge per kilometre does not earn the mark.

12. (a) Second equation × 3: 6x − 3y = 21; adding 4x + 3y = 7: 10x = 28 [1]; x = 14/5 (or 2.8) [1]; y = 2(2.8) − 7, y = −7/5 (or −1.4) [1] (b) 2x − y = 7 gives y = 2x − 7, gradient 2 [1]; kx + 3y = 7 gives y = −(k/3)x + 7/3, so for parallel lines −k/3 = 2 [1]; k = −6: the lines then have the same gradient but different y-intercepts (7/3 and −7), so they are parallel, never meet and there is no solution [1] Examiner insight: The final mark in (b) needs both the value and the reason; k = −6 with no mention of equal gradients and different intercepts does not earn it.

Where marks are usually lost

• Adding the equations when the letter to be removed has the same sign in both.
• Multiplying the left-hand side of an equation by a number but forgetting the right-hand side.
• Sign errors when subtracting a negative term, such as 6y − (−35y).
• Partly expanding a bracket when substituting, such as 2(4x − 5) written as 8x − 5.
• Clearing fractions in only some of the terms of an equation.
• Rounding a fraction answer when the question asks for exact values.
• Stopping after the first value instead of substituting back for the second.
• Answering a context question with “c = 3” instead of “a coffee costs £3.00”, or leaving out units.
• (Higher) Giving x = 3 when the question asks for the coordinates of a point.
• (Higher) Stating that lines are parallel without showing that their gradients are equal.

Next steps

• Refresh the methods: revision notes
• Re-learn anything you missed: study guide
• More Topic 2 practice: algebraic manipulation practice questions
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.6 Simultaneous linear equations: statement 2.6A (Foundation and Higher tiers) and statement 2.6B (Higher tier only).